FOUNDATION · CHAPTER 7 OF 7

How a circuit behaves across frequency

This chapter turns a phasor circuit into a transfer function of frequency. It derives the half-power point of a filter, explains resonance as energy passing between L and C, and defines the quality factor. You need it to predict what a linear circuit does to sinusoids of different frequencies.

Updated 12 Oct 2026

What you will be able to do

You will turn a phasor circuit into a ratio of output to input that depends on frequency, and read its size and phase shift. You will derive the frequency at which a simple filter passes half power. You will explain resonance as energy passing between an inductor and a capacitor, and rate it by a ratio of energies. You will sketch a response from its values at zero, special and very high frequency.

Before you start

Chapter 6, Sinusoids, complex numbers and phasors (steady state of a linear circuit, phasors, impedance, the voltage divider on impedances, average power, rms values)
Energy stored in an inductor, $\frac{1}{2} L i^{2}$ , and in a capacitor, $\frac{1}{2} C v^{2}$
Complex numbers: magnitude, angle, product, quotient, and $1/ j = - j$
Trigonometry: the inverse tangent, $cos (x - 9 0^{\circ}) = sin x$ and $sin^{2} x + cos^{2} x = 1$
Logarithms to base 10
The quadratic formula

From a phasor circuit to H(jω)

Chapter 6 gave you impedance: a ratio of phasors at one frequency. Now keep the circuit fixed and let the frequency vary. The ratio of an output phasor to an input phasor then becomes a function of $ω$ , called the transfer function $H (j ω)$ . Its magnitude says how much the circuit scales a sinusoid; its angle says how far it shifts it. Phasors are RMS and referred to cosine, as before, and $j$ is the imaginary unit.

TRANSFER FUNCTION FROM PHASORS

H (j ω) = \frac{V _{o u t}}{V _{in}} = ∣ H (j ω) ∣ ∠ H (j ω)

Holds for a linear circuit with fixed R, L and C, driven at one $ω$ , once transients have died away (Chapter 6). $V_{in}$ , $V_{o u t}$ : phasors (V, rms); $∣ H ∣$ : a pure ratio; $∠ H$ : rad or degrees. If $v_{in} = 2 ∣ V_{in} ∣ cos (ω t + θ)$ , then $v_{o u t} = 2 ∣ V_{in} ∣∣ H ∣ cos (ω t + θ + ∠ H)$ .

RC LOW-PASS TRANSFER FUNCTION

H (j ω) = \frac{1/ ( j ω C )}{R + 1/ ( j ω C )} = \frac{1}{1 + j ω R C}

A source in series with R and C; the output is the voltage across C and nothing is connected to it, so one current flows through both and the divider applies. Multiply top and bottom by $j ω C$ . R in ohm, C in farad, $ω$ in rad/s.

MAGNITUDE AND PHASE OF THE RC LOW-PASS

∣ H ∣ = \frac{1}{1 + ( ω R C ) ^{2}}, ∠ H = - arctan (ω R C)

A quotient's magnitude is the quotient of the magnitudes; its angle is the difference of the angles. The numerator 1 has angle 0. The denominator $1 + j ω R C$ has magnitude $1 + (ω R C)^{2}$ and, its real part being positive, angle $arctan (ω R C)$ . The output lags the input.

THE HALF-POWER POINT

∣ H ∣ = \frac{1}{2} \Rightarrow 1 + (ω_{c} R C)^{2} = 2 \Rightarrow ω_{c} = \frac{1}{R C}

Defined as the frequency where the gain falls to $1/ 2$ of its value at $ω = 0$ (which is 1). A power set by $V_{o u t}$ in a fixed resistance that does not load the filter scales as $∣ V_{o u t} ∣^{2}$ , so it halves there. $ω_{c}$ in rad/s; $f_{c} = ω_{c} /2 π$ in Hz.

ω	\|H\|	∠H	Meaning
0	1	0°	passed in full
ω_c = 1/RC	0.707	-45°	half power
∞	0	-90°	blocked

Decibels, log axes and reading a response

A response can span a factor of a million in gain and in frequency. A linear axis crushes almost all of it against one edge. On a logarithmic axis equal distances mean equal ratios: 100 to 1000 rad/s (one decade) is as wide as 1000 to 10 000 rad/s. A logarithm also turns the product of cascaded gains (stages that do not load each other) into a sum. The decibel is the logarithmic unit built for this.

DECIBEL

G_{dB} = 10 lo g_{10} \frac{P _{2}}{P _{1}}, G_{dB} = 20 lo g_{10} ∣ H ∣

The first holds for any power ratio: $P_{1}$ , $P_{2}$ are input and output powers in W. The second defines the dB gain of an amplitude ratio. Power in R is $∣ V ∣^{2} / R$ , so the two agree only when both voltages sit across equal resistances. Each factor of 10 in $∣ H ∣$ is 20 dB, and $20 lo g_{10} (1/ 2) = - 3.01$ dB: hence the $- 3$ dB point.

WORKED EXAMPLE

A 10 V rms source is in series with R = 1 kΩ and C = 1 µF; the output is the voltage across C. Find the half-power frequency, then the output amplitude, gain in dB and phase at ω = 1000 rad/s and at ω = 10 000 rad/s.

Solution

$R C = 1000 Ω \times 1 0^{- 6} F = 1 0^{- 3}$ s, so $ω_{c} = 1/ R C = 1000$ rad/s. This is where $ω R C = 1$ .
At $ω = 1000$ : $∣ H ∣ = 1/ 2 = 0.707$ and $∠ H = - 4 5^{\circ}$ . The output is $10 \times 0.707 = 7.07$ V rms, lagging the input by $4 5^{\circ}$ , and the gain is $20 lo g_{10} (0.707) = - 3.01$ dB.
At $ω = 10000$ : $ω R C = 10$ , so $∣ H ∣ = 1/ 101 = 0.0995$ and the output is $10 \times 0.0995 = 0.995$ V rms.
The gain is $20 lo g_{10} (0.0995) = - 20.04$ dB and the angle is $- arctan 10 = - 84. 3^{\circ}$ . For $ω ≫ ω_{c}$ , $∣ H ∣ \approx 1/ (ω R C)$ , so each further decade costs about 20 dB.

Answer: $ω_{c} = 1000$ rad/s. At 1000 rad/s: 7.07 V rms, $- 3.01$ dB, $- 4 5^{\circ}$ . At 10 000 rad/s: 0.995 V rms, $- 20.04$ dB, $- 84. 3^{\circ}$ .

You can often skip the algebra and read a response from its end values, using each element's limiting form.

Element	Impedance	At ω = 0	At ω → ∞
Capacitor	1/(jωC)	open circuit	short circuit
Inductor	jωL	short circuit	open circuit

Replace each element by its ω = 0 form and find the output over the input.
Do the same for ω → ∞.
At the special frequency (ω_c or ω₀), put that ω into the impedances and use the divider.
Name the response by what it passes: high at 0 and low at ∞ is low-pass; low at 0 and high at ∞ is high-pass; low at both ends and high at ω₀ is band-pass.

WATCH OUT FOR

Using 10 log on a voltage ratio. Test: an amplitude ratio (voltage, current or $∣ H ∣$ ) takes 20; a power ratio takes 10.

Resonance and Q

Take a series loop of a source, R, L and C. Take each element's voltage with the current entering its + terminal (passive sign convention), so $V_{L} = j ω L I$ and $V_{C} = I / (j ω C) = - j I / (ω C)$ , using $1/ j = - j$ . These have opposite signs, so at one frequency they cancel. Lower-case $i$ , $v$ , $w$ are values at an instant; upper-case are rms phasors. At resonance the energy leaving the inductor enters the capacitor, so the source supplies only what R dissipates.

SERIES RESONANCE

Z = R + j (ω L - \frac{1}{ω C}), ω_{0} = \frac{1}{L C}

Series elements: impedances add. The bracket, the reactance X, is zero when $ω_{0}^{2} L C = 1$ ; then $Z = R$ . For a source phasor V of fixed size, $∣ Z ∣ \geq R$ at every $ω$ , so the current $V / Z$ is largest at $ω_{0}$ : $V / R$ , in phase with V. $ω_{0}$ in rad/s, L in henry, C in farad.

ENERGY EXCHANGE AND Q

w_{L} w_{C} w_{L} + w_{C} Q = \frac{1}{2} L i^{2} = L I^{2} cos^{2} ω_{0} t = \frac{1}{2} C v_{C}^{2} = \frac{I ^{2}}{ω _{0}^{2} C} sin^{2} ω_{0} t = L I^{2} sin^{2} ω_{0} t = L I^{2} = 2 π \frac{L I ^{2}}{I ^{2} R ( 2 π / ω _{0} )} = \frac{ω _{0} L}{R} = \frac{1}{ω _{0} R C}

At $ω_{0}$ , with $i = 2 I cos ω_{0} t$ , I in A rms. The capacitor voltage lags the current by $9 0^{\circ}$ : $v_{C} = \frac{2 I}{ω _{0} C} sin ω_{0} t$ ; then $1/ (ω_{0}^{2} C) = L$ . Q (quality factor, not reactive power) is $2 π \times$ energy stored $\div$ energy R dissipates in one period $2 π / ω_{0}$ s, at average power $I^{2} R$ . w in J.

HALF-POWER BAND AND BANDWIDTH

R^{2} + X^{2} = 2 R^{2} ω^{2} \pm \frac{R}{L} ω - ω_{0}^{2} ω_{1, 2} B \Rightarrow ω L - \frac{1}{ω C} = \mp R = 0 = \mp \frac{R}{2 L} + (\frac{R}{2 L})^{2} + ω_{0}^{2} = ω_{2} - ω_{1} = \frac{R}{L} = \frac{ω _{0}}{Q}

For a fixed source amplitude, current falls to $1/ 2$ of its peak, and power in R halves, where $∣ Z ∣ = 2 R$ . Upper signs: below $ω_{0}$ , $X < 0$ , giving $ω_{1}$ ; lower signs give $ω_{2}$ . Multiply by $ω / L$ . Each quadratic's roots multiply to $- ω_{0}^{2}$ , so keep its one positive root. B in rad/s.

The share of the source across R is $V_{R} / V = R / (R + j X)$ , with angle $- arctan (X / R)$ . X tends to $- \infty$ as $ω \to 0$ and to $+ \infty$ as $ω \to \infty$ , so the angle tends to $+ 9 0^{\circ}$ and $- 9 0^{\circ}$ . These are limits: the magnitude itself goes to 0. The response is band-pass.

ω	Capacitor and inductor act as	\|V_R / V\|	Phase of V_R
→ 0	capacitor open	→ 0	→ +90°
ω₀	cancel each other	1	0°
→ ∞	inductor open	→ 0	→ -90°

WATCH OUT FOR

Thinking resonance means zero voltage across L and C. Test: at $ω_{0}$ each has size $Q ∣ V ∣$ and can exceed the source; only their phasor sum is zero.

WORKED EXAMPLE

A 10 V rms source drives R = 5 Ω, L = 10 mH and C = 100 µF in series. Find $ω_{0}$ , $Q$ , the bandwidth, the half-power frequencies and the rms capacitor voltage at resonance.

Solution

$ω_{0} = 1/ L C = 1/ 1 0^{- 2} \times 1 0^{- 4} = 1000$ rad/s, because the bracket in Z vanishes there.
$Q = ω_{0} L / R = 1000 \times 0.01/5 = 2$ . Check with the other form: $1/ (ω_{0} R C) = 1/ (1000 \times 5 \times 1 0^{- 4}) = 2$ . They agree.
$B = R / L = 5/0.01 = 500$ rad/s, which equals $ω_{0} / Q$ .
$R /2 L = 250$ rad/s and $25 0^{2} + 1 0^{6} = 1030.8$ rad/s, so $ω_{1} = 780.8$ and $ω_{2} = 1280.8$ rad/s. Their difference is 500 rad/s, as B requires.
At $ω_{0}$ , $Z = R$ , so $I = 10/5 = 2$ A. Then $∣ V_{C} ∣ = I / (ω_{0} C) = 2/ (1000 \times 1 0^{- 4}) = 20$ V, which is $Q \times 10$ V.

Answer: $ω_{0} = 1000$ rad/s, $Q = 2$ , $B = 500$ rad/s, half-power at 780.8 and 1280.8 rad/s, and $∣ V_{C} ∣ = 20$ V rms (twice the source).

Check yourself

CHECK YOURSELF 1

An RC low-pass filter (output across C, nothing connected to it) is driven at its half-power frequency $ω_{c} = 1/ R C$ . Compared with its value at very low frequency, $∣ V_{o u t} ∣^{2}$ (to which the power the output sets in any fixed resistance is proportional) is:

(A)
one quarter
(B)
one half
(C)
0.707 times
(D)
the same

Answer and solution

Answer: (B)

At $ω_{c}$ , $∣ H ∣ = 1/ 2$ , and at very low frequency $∣ H ∣ = 1$ .
$∣ V_{o u t} ∣^{2}$ scales by $∣ H ∣^{2} = 1/2$ .
The answer is one half. 0.707 is the amplitude ratio, not the power ratio.

CHECK YOURSELF 2

An RC low-pass filter has R = 2 kΩ and C = 0.5 µF, with the output across C and nothing connected to it. Find its gain $20 lo g_{10} ∣ H ∣$ in dB at $ω = 3000$ rad/s.

Answer and solution

Answer: $- 10$ (accepted range $\pm 0.05$ )

$R C = 2000 \times 0.5 \times 1 0^{- 6} = 1 0^{- 3}$ s, so $ω R C = 3$ .
$∣ H ∣ = 1/ 1 + 9 = 1/ 10 = 0.3162$ .
$G = 20 lo g_{10} (0.3162) = - 10.0$ dB.

CHECK YOURSELF 3

A series R, L, C circuit has R = 20 Ω, L = 40 mH and resonant frequency $ω_{0} = 2000$ rad/s. Find its quality factor Q.

Answer and solution

Answer: $4$ (accepted range $\pm 0.01$ )

$Q = ω_{0} L / R$ .
$Q = 2000 \times 0.04/20 = 80/20 = 4$ .

How a circuit behaves across frequency

What you will be able to do

Before you start

From a phasor circuit to H(jω)

Decibels, log axes and reading a response

Resonance and Q

Check yourself

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